Parametrization of solutions to the Pythagorean equation
1. Given a solution (a, b, c) to x2+y2 =z2 we can assumea and b are relatively prime: to see this let d= (x, y). Then
a
d 2
+
b
d
2
= c d
2
a21+b21 = c21
and (a1, b1) = 1.
2. If a2 +b2 = c2 with (a, b) = 1 we say (a, b, c) is a fundamental solution.
3. If (a, b, c) is a fundamental solution tox2+y2 =z2 then exactly one of a, bis even.
Proof: If both are even then 2 | (a, b) = 1 which is false. If both are odd
a = 2n+ 1, b= 2m+ 1
a2 = 4(n2+n) + 1, b2 = 4(m2+m) + 1
so z2 ≡ 2 mod 4, which is impossible. Therefore one is even and the other odd.
4. If (a, b, c) is a fundamental solution then cis odd.
Proof: Ifa is even and b odd thena2+b2 ≡1 mod 2, soc2 is odd and thus cis odd.
5. Let (a, b, c) be a fundamental solution and suppose that a is even.
Then, we claim there are positive integersp, q with (p, q) = 1, with one even and the other odd, such that
a = 2pq b = p2−q2 c = p2+q2.
We have
a= 2r so a2 = 4r2 =c2−b2 = (c+b)(c−b) (1)
1
2
Now b and c are both odd so c+b and c−b are both even, say c+b = 2s
c−b = 2t.
adding then subtracting these equations gives c=s+t, b=s−t, so by (1),
4r2 = 2s2t =⇒ r2 =st(2)
6. We claim (s, t) = 1: If d| (s, t) then d |c and d| b, but this means d= 1 since (a, b, c) is fundamental.
7. By (2)s andt must both be squares, so we can writes =p2, t=q2 for some positive integers p, q. But then
a2 = 4r2 = 4st= 4p2q2 =⇒ a= 2pq c = s+t =p2−q2
b = s−t=p2−q2 (3).
8. Finally if d | (p, q) then d | (p2, q2) = (s, t) which are coprime so d= 1 and we have p, q coprime. If pand q are both even or both odd, we would have, by (3), a and b both even, which is impossible since one is even and one is odd. This completes the proof.
